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2 years ago

Why the weight of a body decreases with increases in distance from the earth's surface

Physics

2 answers:

Tom [10]2 years ago

6 0

because as the distance increases the gravitational force decreases so the weight of a body decreases

Bess [88]2 years ago

5 0

Answer:

Weight of the body decreases with increase in distance from the Earth because weight is inversely proportional to distance from the Earth which means Weight of the body decreases with increase in distance from the Earth and vice versa. Also Weight is "gravitational force exerted on the body" so as the distance from the Earth increases , the "gravitational force" decreases.

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A slab of glass has a 0.600 cm thick layer of water on top of it. A light ray strikes the water at an incident angle of 59.0°. A

Dimas [21]

Answer:

49.63 degree

Explanation:

thickness of glass slab, t = 0.6 cm

angle of incidence = 59 degree

Let r be the angle of refraction

The refractive index of glass, ng = 3/2

refractive index of water, nw = 4/3

refarctive index of glass with respect to water = ng / nw = 3 /2 ÷ 4 /3 = 9 / 8

So, by use of Snell's law

Refractive index of glass with respect to water = Sin i / Sin r

9 / 8 = Sin 59 / Sin r

9 / 8 = 0.857 / Sin r

Sin r = 0.7619

r = 49.63 degree

4 0

3 years ago

1 point

Sav [38]

Answer:

388.5J

Explanation:

Given parameters:

Weight = 70N

Height = 5.55m

Unknown:

Gravitational potential energy at the top of the ladder = ?

Solution:

The gravitational potential energy is the energy due to the position of the body.

Gravitational potential energy = Weight x height

So;

Gravitational potential energy = 70 x 5.55 = 388.5J

8 0

2 years ago

A force in the +x-direction with magnitude ????(x) = 18.0 N − (0.530 N/m)x is applied to a 6.00 kg box that is sitting on the ho

fiasKO [112]

Answer:

$v_f=8.17\frac{m}{s}$

Explanation:

First, we calculate the work done by this force after the box traveled 14 m, which is given by:

$W=\int\limits^{x_f}_{x_0} {F(x)} \, dx \\W=\int\limits^{14}_{0} ({18N-0.530\frac{N}{m}x}) \, dx\\W=[(18N)x-(0.530\frac{N}{m})\frac{x^2}{2}]^{14}_{0}\\W=(18N)14m-(0.530\frac{N}{m})\frac{(14m)^2}{2}-(18N)0+(0.530\frac{N}{m})\frac{0^2}{2}\\W=252N\cdot m-52N\cdot m\\W=200N\cdot m$

Since we have a frictionless surface, according to the the work–energy principle, the work done by all forces acting on a particle equals the change in the kinetic energy of the particle, that is:

$W=\Delta K\\W=K_f-K_i\\W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}$